Question

An unknown amount of a compound with a molecular mass of 260.73 g/mol is dissolved in a 10-mL volumetric flask. A 1.00-mL aliquot of this solution is transferred to a 25-mL volumetric flask and enough water is added to dilute to the mark. The absorbance of this diluted solution at 353 nm is 0.565 in a 1.000-cm cuvette. The molar absorptivity for this compound at 353 nm is ε353 = 5793 M–1 cm–1.
(a) What is the concentration of the compound in the cuvette?
(b) What is the concentration of the compound in the 10-mL flask?
(c) How many milligrams of compound were used to make the 10-mL solution?

Answer 1

Answer:

a) $9.7532\times 10^{-5} M$ is the concentration of the compound in the cuvette.

b) 0.02438 M is the concentration of the solution in the 10-mL flask.

c) 63.57 milligrams of compound were used to make the 10-mL solution.

Explanation:

Using Beer-Lambert's law :

Formula used :

$A=\epsilon \times c\times l$

where,

A = absorbance of solution

c = concentration of solution =

a)  We have :

$\epsilon =5793 M^{-1} cm^{-1}$

l = path length = 1.000 cm

A = 0.565 , c = ?

$0.565=5793 M^{-1} cm^{-1}\times c\times 1.000 cm$

$c=9.7532\times 10^{-5} M$

$9.7532\times 10^{-5} M$ is the concentration of the compound in the cuvette.

b) The initial concentration of solution in 1 mL = $C_1$

$V_1=1 mL$

Final concentration of solution after dilution = $C_2=9.7532\times 10^{-5} M$

$V_2=25 mL$

$C_1V_1=C_2V_2$

$C_1=\frac{C_2V_2}{V_1}=\frac{9.7532\times 10^{-5} M\times 25 mL}{1 mL}$

$C_1=0.002438 M$

Concentration of 1 mL solution = 0.002438 M

Then concentration of the 10 mL solution :

$10 mL\times 0.002438 M = 0.02438 M$

0.02438 M is the concentration of the compound in the 10-mL flask.

c) Amount of compound were used to make the 10-mL solution =m

Molarity of the solution = 0.02438 M

volume of the solution = 10 m L = 0.010 L

$Molarity=\frac{Moles}{Volume(L)}$

$0.02438 M=\frac{Moles}{0.010 L}$

Moles of compound:

0.02438 M × 0.010 L= 0.0002438 mol

Mass of  0.0002438 mol unknown compound =

0.0002438 mol × 260.73 g/mol = 0.06357 g

0.06357 g = 63.57 mg (1 g = 1000 mg)

63.57 milligrams of compound were used to make the 10-mL solution.